Chapter 25, Problem 1TS

Gold has a density of $19.3\text{g}/{\text{cm}}^{\text{3}}$. A $5\text{-gal}$ pail of water (density of water $=1.0\text{g}/{\text{cm}}^{\text{3}}$) has a mass of about $18\text{kg}$. What is the mass of a $5\text{-gal}$ pail of gold?

To determine

The mass of a $5\text{-gal}$ pail of gold.

Answer to Problem 1TS

The mass of a $5\text{-gal}$ pail of gold is $347.4\text{kg}$.

Explanation of Solution

Given Info: The density of the Gold, $d_g$ is $19.3\text{g}/{\text{cm}}^{\text{3}}$, the density of water $d_w$ is $1.0\text{g}/{\text{cm}}^{\text{3}}$ and the mass of $5\text{-gal}$ pail of water is $m_w$ $18\text{kg}$.

Write the expression to calculate the specific gravity in terms of density.

$S.G.=\frac{d_g}{d_w}$

Here,

$d_g$ is mass of gold

$d_w$ is the mass of water

Write the expression to calculate the specific gravity in terms of mass.

$S.G=\frac{m_g}{m_w}$

Here,

$m_g$ is mass of gold

$m_w$ is the mass of water

Equate RHS of both the expression.

$\begin{array}{c}\frac{d_g}{d_w}=\frac{m_g}{m_w}\\ m_w{d}_g=m_g{d}_w\\ m_g=\frac{m_w{d}_g}{d_w}\end{array}$

Substitute $19.3\text{g}/{\text{cm}}^{\text{3}}$ for $d_g$, $1.0\text{g}/{\text{cm}}^{\text{3}}$ for $d_w$, and $18\text{kg}$ for $m_w$ in the above expression to get $m_g$.

$\begin{array}{c}{m}_g\text{=}\frac{19.3{\text{ g/cm}}^3}{1{\text{g/cm}}^3}\times 18\text{kg}\\ =347.4\text{kg}\end{array}$

Conclusion:

Therefore, the mass of a $5-\text{gal}$ pail of gold is $347.4\text{kg}$.